TITLE.PM5

430 ENGINEERING THERMODYNAMICS

dharm

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Now cvCO 2 =

RCO 2

γ− 1

=

8 314

44 1 286 1

.

(. − ) [forCO.]Qγ^2 =1 286

= 0.661 kJ/kg K

and cpCO 2 = 1.286 × 0.661 = 0.85 kJ/kg K

For the mixture :

cp =

mc m c

mm

NCp Op

NCO

2 NC 2 2 O 2

22

+

+

=

4 1 039 6 0 85

46

×+×

+

..

() = 0.925 kJ/kg K. (Ans.)

cv =

mc m c

mm

NCvvO

NCO

2 NC 2 2 O 2

22

+

+

=

4 0 742 6 0 661

46

×+×

+

..

() = 0.693 kJ/kg K. (Ans.)

When the mixture is heated at constant volume :

Change in internal energy,

U 2 – U 1 = mcv(T 2 – T 1 ) = 10 × 0.693(50 – 20) = 207.9 kJ. (Ans.)

Change in entropy,

H 2 – H 1 = mcp(T 2 – T 1 ) = 10 × 0.925(50 – 20) = 277.5 kJ. (Ans.)

Change in entropy,

S 2 – S 1 = mcv loge T

T

2

1

+ mR loge V

V

2

1

= mcv loge

T

T

2

1

()QVV 12 =

= 10 × 0.693 × loge

273 50

273 20

+

+

F

HG

I

KJ

= 0.675 kJ/kg K. (Ans.)

When the mixture is heated at constant pressure :

If the mixture is heated at constant pressure ∆U and ∆H will remain the same.

The change in entropy will be

S 2 – S 1 = mcp loge

T

T

2

1

• mR loge

p

p

2

1

= mcp loge

T

T

2

1

()Qpp 12 =

= 10 × 0.925 loge

323

293

F

HG

I
KJ = 0.902 kJ/kg K. (Ans.)
Example 9.11. A vessel of 1.8 m^3 capacity contains oxygen at 8 bar and 50°C. The vessel is
connected to another vessel of 3.6 m^3 capacity containing carbon monoxide at 1 bar and 20°C. A
connecting valve is opened and the gases mix adiabatically. Calculate :
(i)The final temperature and pressure of the mixture ;
(ii)The change of entropy of the system.
Take : For oxygen Cv = 21.07 kJ/mole K.
For carbon monoxide Cv = 20.86 kJ/mole K.